You have found the following ages (in years) of all 4 zebras at your local zoo: $ 18,\enspace 5,\enspace 13,\enspace 8$ What is the average age of the zebras at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 4 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{18 + 5 + 13 + 8}{{4}} = {11\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $18$ years $7$ years $49$ years $^2$ $5$ years $-6$ years $36$ years $^2$ $13$ years $2$ years $4$ years $^2$ $8$ years $-3$ years $9$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{49} + {36} + {4} + {9}} {{4}} $ $ {\sigma^2} = \dfrac{{98}}{{4}} = {24.5\text{ years}^2} $ The average zebra at the zoo is 11 years old. The population variance is 24.5 years $^2$.